Integrand size = 33, antiderivative size = 36 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {1-d} (1-x)}{\sqrt {-1+x^3}}\right )}{\sqrt {1-d}} \] Output:
-2*arctanh((1-d)^(1/2)*(1-x)/(x^3-1)^(1/2))/(1-d)^(1/2)
Time = 2.61 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {-1+d} \sqrt {-1+x^3}}{1+x+x^2}\right )}{\sqrt {-1+d}} \] Input:
Integrate[(2 + 2*x - x^2)/((2 - d + d*x + x^2)*Sqrt[-1 + x^3]),x]
Output:
(2*ArcTan[(Sqrt[-1 + d]*Sqrt[-1 + x^3])/(1 + x + x^2)])/Sqrt[-1 + d]
Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2570, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-x^2+2 x+2}{\sqrt {x^3-1} \left (d x-d+x^2+2\right )} \, dx\) |
\(\Big \downarrow \) 2570 |
\(\displaystyle 4 \int \frac {1}{\frac {2 (1-d) (1-x)^2}{x^3-1}-2}d\frac {1-x}{\sqrt {x^3-1}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\frac {\sqrt {1-d} (1-x)}{\sqrt {x^3-1}}\right )}{\sqrt {1-d}}\) |
Input:
Int[(2 + 2*x - x^2)/((2 - d + d*x + x^2)*Sqrt[-1 + x^3]),x]
Output:
(-2*ArcTanh[(Sqrt[1 - d]*(1 - x))/Sqrt[-1 + x^3]])/Sqrt[1 - d]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((f_) + (g_.)*(x_) + (h_.)*(x_)^2)/(((c_) + (d_.)*(x_) + (e_.)*(x_)^2)* Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Simp[-2*g*h Subst[Int[1/(2*e*h - (b*d*f - 2*a*e*h)*x^2), x], x, (1 + 2*h*(x/g))/Sqrt[a + b*x^3]], x] /; Fre eQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b*d*f - 2*a*e*h, 0] && EqQ[b*g^3 - 8 *a*h^3, 0] && EqQ[g^2 + 2*f*h, 0] && EqQ[b*d*f + b*c*g - 4*a*e*h, 0]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.69 (sec) , antiderivative size = 4437, normalized size of antiderivative = 123.25
method | result | size |
default | \(\text {Expression too large to display}\) | \(4437\) |
elliptic | \(\text {Expression too large to display}\) | \(4646\) |
Input:
int((-x^2+2*x+2)/(d*x+x^2-d+2)/(x^3-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2*(-3/2-1/2*I*3^(1/2))*((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x+1/2-1/2*I*3 ^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2)*((x+1/2+1/2*I*3^(1/2))/(3/2+1/2*I*3^(1/ 2)))^(1/2)/(x^3-1)^(1/2)*EllipticF(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),((3/ 2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))+3/2/(d^2+4*d-8)^(1/2)*(1/(-3/ 2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^(1/2))*x+ 1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*(1/(3/2+1 /2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))*3^(1/2)) ^(1/2)/(x^3-1)^(1/2)/(1+1/2*d-1/2*(d^2+4*d-8)^(1/2))*EllipticPi(((x-1)/(-3 /2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(1+1/2*d-1/2*(d^2+4*d-8)^(1/2 )),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))*d^2+4*I/(d^2+4*d-8)^(1 /2)*(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I* 3^(1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2 )*(1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2 ))*3^(1/2))^(1/2)/(x^3-1)^(1/2)/(1+1/2*d+1/2*(d^2+4*d-8)^(1/2))*EllipticPi (((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),(3/2+1/2*I*3^(1/2))/(1+1/2*d+1/2*(d^2+ 4*d-8)^(1/2)),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))*3^(1/2)-3/2 *(1/(-3/2-1/2*I*3^(1/2))*x-1/(-3/2-1/2*I*3^(1/2)))^(1/2)*(1/(3/2-1/2*I*3^( 1/2))*x+1/2/(3/2-1/2*I*3^(1/2))-1/2*I/(3/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*( 1/(3/2+1/2*I*3^(1/2))*x+1/2/(3/2+1/2*I*3^(1/2))+1/2*I/(3/2+1/2*I*3^(1/2))* 3^(1/2))^(1/2)/(x^3-1)^(1/2)/(1+1/2*d-1/2*(d^2+4*d-8)^(1/2))*EllipticPi...
Time = 0.12 (sec) , antiderivative size = 187, normalized size of antiderivative = 5.19 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=\left [-\frac {\sqrt {-d + 1} \log \left (-\frac {2 \, {\left (3 \, d - 4\right )} x^{3} - x^{4} - {\left (d^{2} - 2 \, d + 4\right )} x^{2} - d^{2} + 4 \, \sqrt {x^{3} - 1} {\left ({\left (d - 2\right )} x - x^{2} - d\right )} \sqrt {-d + 1} + 2 \, {\left (d^{2} - 2 \, d\right )} x - 4 \, d + 4}{2 \, d x^{3} + x^{4} + {\left (d^{2} - 2 \, d + 4\right )} x^{2} + d^{2} - 2 \, {\left (d^{2} - 2 \, d\right )} x - 4 \, d + 4}\right )}{2 \, {\left (d - 1\right )}}, -\frac {\arctan \left (-\frac {\sqrt {x^{3} - 1} {\left ({\left (d - 2\right )} x - x^{2} - d\right )} \sqrt {d - 1}}{2 \, {\left ({\left (d - 1\right )} x^{3} - d + 1\right )}}\right )}{\sqrt {d - 1}}\right ] \] Input:
integrate((-x^2+2*x+2)/(d*x+x^2-d+2)/(x^3-1)^(1/2),x, algorithm="fricas")
Output:
[-1/2*sqrt(-d + 1)*log(-(2*(3*d - 4)*x^3 - x^4 - (d^2 - 2*d + 4)*x^2 - d^2 + 4*sqrt(x^3 - 1)*((d - 2)*x - x^2 - d)*sqrt(-d + 1) + 2*(d^2 - 2*d)*x - 4*d + 4)/(2*d*x^3 + x^4 + (d^2 - 2*d + 4)*x^2 + d^2 - 2*(d^2 - 2*d)*x - 4* d + 4))/(d - 1), -arctan(-1/2*sqrt(x^3 - 1)*((d - 2)*x - x^2 - d)*sqrt(d - 1)/((d - 1)*x^3 - d + 1))/sqrt(d - 1)]
\[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=- \int \left (- \frac {2 x}{d x \sqrt {x^{3} - 1} - d \sqrt {x^{3} - 1} + x^{2} \sqrt {x^{3} - 1} + 2 \sqrt {x^{3} - 1}}\right )\, dx - \int \frac {x^{2}}{d x \sqrt {x^{3} - 1} - d \sqrt {x^{3} - 1} + x^{2} \sqrt {x^{3} - 1} + 2 \sqrt {x^{3} - 1}}\, dx - \int \left (- \frac {2}{d x \sqrt {x^{3} - 1} - d \sqrt {x^{3} - 1} + x^{2} \sqrt {x^{3} - 1} + 2 \sqrt {x^{3} - 1}}\right )\, dx \] Input:
integrate((-x**2+2*x+2)/(d*x+x**2-d+2)/(x**3-1)**(1/2),x)
Output:
-Integral(-2*x/(d*x*sqrt(x**3 - 1) - d*sqrt(x**3 - 1) + x**2*sqrt(x**3 - 1 ) + 2*sqrt(x**3 - 1)), x) - Integral(x**2/(d*x*sqrt(x**3 - 1) - d*sqrt(x** 3 - 1) + x**2*sqrt(x**3 - 1) + 2*sqrt(x**3 - 1)), x) - Integral(-2/(d*x*sq rt(x**3 - 1) - d*sqrt(x**3 - 1) + x**2*sqrt(x**3 - 1) + 2*sqrt(x**3 - 1)), x)
Exception generated. \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((-x^2+2*x+2)/(d*x+x^2-d+2)/(x^3-1)^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d^2-4*(2-d)>0)', see `assume?` f or more de
\[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=\int { -\frac {x^{2} - 2 \, x - 2}{\sqrt {x^{3} - 1} {\left (d x + x^{2} - d + 2\right )}} \,d x } \] Input:
integrate((-x^2+2*x+2)/(d*x+x^2-d+2)/(x^3-1)^(1/2),x, algorithm="giac")
Output:
integrate(-(x^2 - 2*x - 2)/(sqrt(x^3 - 1)*(d*x + x^2 - d + 2)), x)
Time = 0.11 (sec) , antiderivative size = 629, normalized size of antiderivative = 17.47 \[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=\text {Too large to display} \] Input:
int((2*x - x^2 + 2)/((x^3 - 1)^(1/2)*(d*x - d + x^2 + 2)),x)
Output:
(2*((3^(1/2)*1i)/2 + 3/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3 /2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticF(asin((-(x - 1)/((3^(1/2)*1i)/ 2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(((3^(1 /2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1 /2)*1i)/2 + 1/2) + 1) + x^3)^(1/2) + (2*((3^(1/2)*1i)/2 + 3/2)*(-(x - (3^( 1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2 )/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*el lipticPi(((3^(1/2)*1i)/2 + 3/2)/(d/2 - (4*d + d^2 - 8)^(1/2)/2 + 1), asin( (-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2 )*1i)/2 - 3/2))*(d + (d + 2)*(d/2 - (4*d + d^2 - 8)^(1/2)/2) - 4))/((((3^( 1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^( 1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2)*(d/2 - (4*d + d^2 - 8)^(1/2)/2 + 1)*(4 *d + d^2 - 8)^(1/2)) - (2*((3^(1/2)*1i)/2 + 3/2)*(-(x - (3^(1/2)*1i)/2 + 1 /2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i )/2 + 3/2))^(1/2)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticPi(((3^( 1/2)*1i)/2 + 3/2)/(d/2 + (4*d + d^2 - 8)^(1/2)/2 + 1), asin((-(x - 1)/((3^ (1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2) )*(d + (d + 2)*(d/2 + (4*d + d^2 - 8)^(1/2)/2) - 4))/((((3^(1/2)*1i)/2 - 1 /2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 ...
\[ \int \frac {2+2 x-x^2}{\left (2-d+d x+x^2\right ) \sqrt {-1+x^3}} \, dx=2 \left (\int \frac {\sqrt {x^{3}-1}}{d \,x^{4}+x^{5}-d \,x^{3}+2 x^{3}-d x -x^{2}+d -2}d x \right )-\left (\int \frac {\sqrt {x^{3}-1}\, x^{2}}{d \,x^{4}+x^{5}-d \,x^{3}+2 x^{3}-d x -x^{2}+d -2}d x \right )+2 \left (\int \frac {\sqrt {x^{3}-1}\, x}{d \,x^{4}+x^{5}-d \,x^{3}+2 x^{3}-d x -x^{2}+d -2}d x \right ) \] Input:
int((-x^2+2*x+2)/(d*x+x^2-d+2)/(x^3-1)^(1/2),x)
Output:
2*int(sqrt(x**3 - 1)/(d*x**4 - d*x**3 - d*x + d + x**5 + 2*x**3 - x**2 - 2 ),x) - int((sqrt(x**3 - 1)*x**2)/(d*x**4 - d*x**3 - d*x + d + x**5 + 2*x** 3 - x**2 - 2),x) + 2*int((sqrt(x**3 - 1)*x)/(d*x**4 - d*x**3 - d*x + d + x **5 + 2*x**3 - x**2 - 2),x)